Câu hỏi 39

\(\begin{array}{l}\int\limits_1^2 {{{\left( {\frac{{x - 1}}{{x + 2}}} \right)}^2}dx}  = \int\limits_1^2 {{{\left( {1 - \frac{3}{{x + 2}}} \right)}^2}dx}  = \int\limits_1^2 {\left( {1 - \frac{6}{{x + 2}} + \frac{9}{{{{\left( {x + 2} \right)}^2}}}} \right)dx} \\ = \left. {\left( {x - 6\ln \left| {x + 2} \right| - \frac{9}{{x + 2}}} \right)} \right|_1^2 = 2 - 6\ln 4 - \frac{9}{4} - 1 + 6\ln 3 + 3\\ = 6\ln \frac{3}{4} + \frac{7}{4}\end{array}\)

Chọn D.